A set P = {1, 2, 3, 4, 5} and set Q = {1, 2, 3, 4, 5, 6, 7} is defined with given numbers. Now one number is chosen randomly from each of the given set in such a way that the absolute difference between the two numbers is 2. What is the probability that one of the number chosen is 3 ?
Please post the exact wording of the original question along with the source. On this particular question we can determine what's being asked, but grammatical errors (of which there are a few in this post) can alter the meaning of a question. Also, please post the answers, since many GMAT strategies revolve around using the choices.arjunshn wrote:A set P = {1, 2, 3, 4, 5} and set Q = {1, 2, 3, 4, 5, 6, 7} is defined with given numbers. Now one number is chosen randomly from each of the given set in such a way that the absolute difference between the two numbers is 2. What is the probability that one of the number chosen is 3 ?
Let's start solving this question by noting the probability formula (on test day you should always jot applicable formulae down on your noteboard):
Probability = (# of desired outcomes)/(total # of possibilities)
The denominator is the easier part of the fraction to find, so let's start by calculating the total number of ways we can get two numbers that are 2 apart. I'm sure there are fancy methods, but brute force is probably the quickest.
Focusing on set P, since that's smaller:
1 in P goes with 3 in Q.
2 in P goes with 4 in Q.
3 in P goes with 1 or 5 in Q.
4 in P goes with 2 or 6 in Q.
5 in P goes with 3 or 7 in Q.
So, there are 8 pairs we can select that will have an absolute difference of 2.
For the number of desired outcomes, we now simply count the possibilities than include 3:
1 in P with 3 in Q.
3 in P with 1 or 5 in Q.
5 in P with 3 in Q.
So, there are 4 pairs that match our desired outcome.
Finally, we plug into the equation:
Probability = 4/8 = 1/2... all done!
